Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $a = \dfrac{5z^2 + 25z}{z - 7} \times \dfrac{-2z^2 + 2}{z^3 + 6z^2 + 5z} $
Solution: First factor out any common factors. $a = \dfrac{5z(z + 5)}{z - 7} \times \dfrac{-2(z^2 - 1)}{z(z^2 + 6z + 5)} $ Then factor the quadratic expressions. $a = \dfrac {5z(z + 5)} {z - 7} \times \dfrac {-2(z + 1)(z - 1)} {z(z + 1)(z + 5)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {5z(z + 5) \times -2(z + 1)(z - 1) } {(z - 7) \times z(z + 1)(z + 5) } $ $a = \dfrac {-10z(z + 1)(z - 1)(z + 5)} {z(z + 1)(z + 5)(z - 7)} $ Notice that $(z + 1)$ and $(z + 5)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {-10z\cancel{(z + 1)}(z - 1)(z + 5)} {z\cancel{(z + 1)}(z + 5)(z - 7)} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $a = \dfrac {-10z\cancel{(z + 1)}(z - 1)\cancel{(z + 5)}} {z\cancel{(z + 1)}\cancel{(z + 5)}(z - 7)} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $a = \dfrac {-10z(z - 1)} {z(z - 7)} $ $ a = \dfrac{-10(z - 1)}{z - 7}; z \neq -1; z \neq -5 $